We guess, based on historical data, that 30% of graduating high-school seniors in a large city will have completed a first-year calculus course. What's the minimum sample size needed to construct a 95% confidence interval for a proportion with a margin of error of 2.5%?
Try this formula:
n = [(z-value)^2 * p * (1-p)]/E^2
...where n = sample size you need, z-value = 1.96 to represent 95% confidence, p and 1-p represent proportions, E = .025 (or 2.5%), and ^ means squared.
With your data:
n = [(1.96)^2 * .30 * .70]/.025^2
I'll let you finish the calculation. Round the answer to the next highest whole number.
To find the minimum sample size needed to construct a confidence interval for a proportion, we need to use the formula:
n = (Z^2 * p * (1 - p)) / E^2
where:
- n is the minimum sample size needed
- Z is the z-score corresponding to the desired confidence level (in this case, 95%, which corresponds to a z-score of approximately 1.96)
- p is the estimated proportion (in this case, 30% or 0.3)
- E is the margin of error (in this case, 2.5% or 0.025)
Substituting these values into the formula, we get:
n = (1.96^2 * 0.3 * (1 - 0.3)) / 0.025^2
Calculating:
n = (3.8416 * 0.3 * 0.7) / 0.000625
n ≈ 663.96
Since we cannot have a fractional sample size, rounding up to the nearest whole number, we find that the minimum sample size needed is 664.